fzu-Wand

题目：

N wizards are attending a meeting. Everyone has his own magic wand. N magic wands was put in a line, numbered from 1 to n(Wand_i owned by wizard_i). After the meeting, n wizards will take a wand one by one in the order of 1 to n. A boring wizard decided to reorder the wands. He is wondering how many ways to reorder the wands so that at least k wizards can get his own wand.

For example, n=3. Initially, the wands are w1 w2 w3. After reordering, the wands become w2 w1 w3. So, wizard 1 will take w2, wizard 2 will take w1, wizard 3 will take w3, only wizard 3 get his own wand.

Input
First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: Two number n and k.

1<=n <=10000.1<=k<=100. k<=n.

Output
For each test case, output the answer mod 1000000007(10^9 + 7).

Sample Input
2
1 1
3 1

Sample Output
1
4

题目大意：

给你n个物品和一个数k，要你将其进行重新排列，使得这n个物品中至少存在k个物品是在自己原来的位置上的。

题目思路：

错排+排列组合

注意求组合数时要用逆元，这里用的费马小定理。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1e9+7;const int INF = 0x3f3f3f3f;const int maxn = 12345;int T,n,m;ll D[maxn];ll fac[maxn];ll ans[maxn];void Get_D(){    D[0]=1;    D[1]=0;    D[2]=1;    for(int i=3;i<=10000;i++)        D[i]=(((i-1)%mod)*((D[i-1]+D[i-2])%mod))%mod;}void Get_fac(){    fac[0]=1;    for(int i=1;i<=10000;i++)        fac[i]=(fac[i-1]*i)%mod;}ll Pow_q(ll a,ll b){    ll ans=1;    while(b)    {        if(b&1)            ans=(ans*a)%mod;        a=(a*a)%mod;        b>>=1;    }    return ans%mod;}ll Get_C(ll a,ll b){    ll aa,bb;    bb=fac[b];    aa=(fac[b-a]*fac[a])%mod;    ll aa_ni=Pow_q(aa,mod-2);    return ((bb%mod)*(aa_ni%mod))%mod;}int main(){    scanf("%d",&T);    Get_D();    Get_fac();    while(T--)    {        scanf("%d%d",&n,&m);        ll ans=0;        for(int i=m;i<=n;i++)        {            ll c=Get_C(n-i,n)%mod;            ll d=D[n-i]%mod;            ans=(ans+(c*d)%mod)%mod;        }        printf("%d\n",ans);    }    return 0;}